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Fundamental Concepts Test - 5

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Fundamental Concepts Test - 5
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Find and correct errors of the followingmathematical expressions: 

    7z+55=7z
    Solution

    we will get thecorrect answer by separating the denominators, which is 7z+55=1+7z5

  • Question 2
    1 / -0

    Find product of following pairs of monomial 4h,7h

    Solution
    -4h×7h=-28h2
  • Question 3
    1 / -0

    Find and correct errors of the followingmathematical expressions: 

    (2z+3b)(z-b)=2z2-3b2
    Solution

    Middle term ismissing in the expression correct statement is (2z+3b)(z-b)=2z2+zb-3b2

  • Question 4
    1 / -0

    Carry out the following division 66mn2o3÷11no2

    Solution

    66mn2o311no26mnoAs we know 66/11 is 6and we substract the powers of the same variables which means n2-1= n and o3-2=0

    So the answer aftersimplification is 6mno.

  • Question 5
    1 / -0

    Factorise the expression and divide them asdirected 4az(z2+6z162a(z+8)

    Solution

    4az(z2+6z-16)2a(z+8)

    2zz(z+8)-2(z+8)z+8= 2z (z-2)(z+8)(z+8)

    both the (z+8) willbe cancelled out and we’ll be left with 2z(z-2)

  • Question 6
    1 / -0

    Subtract 4m-7mb+3b+12 from 12m-9mb+5b-3

    Solution

    12m-9mb+5b-3-(4m-7mb+3b+12)

    12m-9mb+5b-3-4m+7mb-3b-12

    8m+2b-2mb-15

  • Question 7
    1 / -0

    Divide the polynomial by the given monomial (5z2-6z)÷3z

    Solution

    5z2-6z3z=5z23z-6z3z

    5z3-2=135z-6
  • Question 8
    1 / -0

    Divide as directed. 5(2g+1)(3g+5)÷(2g+1)

    Solution

    5(2g+1)(3g+5)(2g+1)

    Both (2g+1) will be cancelled outand we’ll be left with 5(3g+5).

  • Question 9
    1 / -0

    Obtain the product of m,2n,3o,6mno

    Solution

    m×2n×3o×6mno

    =36m2n2o2

  • Question 10
    1 / -0

    Find areas of rectangles with following pairs ofmonomials as their length and breadth respectively. 

    20j2,5k2
    Solution

    Area of rectangle,A=Length×Breadth

    20j2×5k2

    =100j2k2
  • Question 11
    1 / -0

    Simplify 3e(4e5)+33e(4e−5)+3 and find itsvalue for e = 3

    Solution

    3e(4e-5)+3
    3×3(4×3-5)+3
    9(7)+3=66

  • Question 12
    1 / -0

    Find and correct errors of the followingmathematical expressions: 3z3z+2=12

    Solution

    As we cannot separate the denominators, the statement will remain as itis the correct statement is 3z3z+2=33z+2

  • Question 13
    1 / -0

    Add a-x+ax,x-c+xc,c-a+ac

    Solution

    a-x+ax,x-c+xc,c-a+ac

    Adding this will give

    a-x+ax+x-c+xc+c-a+ac

    =ax+xc+ac
  • Question 14
    1 / -0

    the following division (10g-25)÷(2g-5)

    Solution

    (10g-25)(2g-5)=5(2g-5)(2g-5)

    Both (2g-5) will be cancelled out and theanswer will be 5

  • Question 15
    1 / -0

    Obtain the volume of rectangular box withfollowing length, breadth and height respectively. 

    z,2z2,3z4
    Solution

    Volume=Length × Breadth × Height

    V=l × b × h

    V= z×2z2×3z4=6z7

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